If sin 2 θ and cos 2 θ are solutions of x2+b x-c=0, then

Question

If sin 2 θ and cos 2 θ are solutions of x2+b x-c=0, then (A) b2+2 c+1=0 (B) b2+2 c-1=0 (C) b2-2 c+1=0 (D) b2-2 c-1=0

Tardigrade Admin · Accepted Answer

According to given informations sin 2 θ+ cos 2 θ=-b and sin 2 θ cos 2 θ=-c ∵ ( sin 2 θ+ cos 2 θ)2= sin 2 2 θ+ cos 2 2 θ+2 sin 2 θ cos 2 θ ⇒ (-b)2=1+2(-c) ⇒ b2+2 c-1=0

Q. If $sin 2 θ$ and $cos 2 θ$ are solutions of $x^{2} + b x - c = 0$ , then

$b^{2} + 2 c + 1 = 0$

$b^{2} + 2 c - 1 = 0$

$b^{2} - 2 c + 1 = 0$

$b^{2} - 2 c - 1 = 0$

According to given informations
$sin 2 θ + cos 2 θ = - b$ and
$sin 2 θ cos 2 θ = - c$
$∵ (sin 2 θ + cos 2 θ)^{2} = sin^{2} 2 θ + cos^{2} 2 θ + 2 sin 2 θ cos 2 θ$
$\Rightarrow (- b)^{2} = 1 + 2 (- c)$
$\Rightarrow b^{2} + 2 c - 1 = 0$

Q. If sin2θ and cos2θ are solutions of x2+bx−c=0, then

b2+2c+1=0

b2+2c−1=0

b2−2c+1=0

b2−2c−1=0